How will we prove that sin^-1[sin(3π/4)] is not equal to 3π/4?

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Answer 1 · 2018-04-18T10:20:31

Please refer to the Explanation.

Recall the following Definition of #sin^-1# function :

#sin^-1x=theta, |x| le 1 hArr sintheta=x, theta in [-pi/2,pi/2]#.

Replacing #x" by "sintheta#, we get,

# sin^-1(sintheta)=theta, theta in [-pi/2,pi/2]#.

Now, #sin^-1(sin(3pi/4))=sin^-1{sin(pi-pi/4)}#

#=sin^-1(sin(pi/4))#,

#=pi/4, because, pi/4 in [-pi/2,pi/2]#.

# rArr sin^-1(sin(3pi/4))=pi/4!=3pi/4#.